Let's solve the given problem step by step based on the information in the diagram:

Given:

1. $AFGB$ is a straight line.
2. $BCEF$ is a parallelogram.
3. Triangle $EFG$ is an isosceles triangle with $EF = EG$.
4. $\angle EAF = 40^\circ$ and $\angle CEF = 65^\circ$.

(i) To find $\angle GFE$:

• Since $EFG$ is an isosceles triangle with $EF = EG$, the base angles $\angle GFE$ and $\angle FEG$ are equal.
• The sum of the angles in a triangle is $180^\circ$, so: $\angle GFE + \angle FEG + \angle EFG = 180^\circ$
• Given that $\angle EFG = 65^\circ$, we can set $\angle GFE = \angle FEG = x$. $x + x + 65^\circ = 180^\circ$ $2x = 115^\circ$ $x = 57.5^\circ$ Therefore, $\angle GFE = 57.5^\circ$.

(ii) To find $\angle FEA$:

• Since $AFGB$ is a straight line, the angles along this line add up to $180^\circ$. Therefore: $\angle FEA + \angle EAF = 180^\circ$ $\angle FEA + 40^\circ = 180^\circ$ $\angle FEA = 140^\circ$

Answers:

1. $\angle GFE = 57.5^\circ$
2. $\angle FEA = 140^\circ$

Would you like more details on this, or do you have any questions?

Here are 5 related questions to consider:

1. What is the measure of $\angle FEG$ in triangle $EFG$?
2. How can the properties of a parallelogram be used to find other angles in the diagram?
3. What is the measure of $\angle BCF$ in the parallelogram $BCEF$?
4. How can the exterior angle theorem help solve problems in triangle geometry?
5. What properties of isosceles triangles are essential in solving angle problems?

Tip: Remember that the sum of angles in a triangle is always $180^\circ$, and this fundamental rule can help solve various geometric problems.